Open Access

Sharp constants for inequalities of Poincaré type: an application of optimal control theory

Journal of Inequalities and Applications20142014:523

https://doi.org/10.1186/1029-242X-2014-523

Received: 24 December 2013

Accepted: 23 April 2014

Published: 29 December 2014

Abstract

Sharp constants for an inequality of Poincaré type are studied. The problem is solved by using optimal control theory.

MSC:26D10, 46E35, 49K15.

Keywords

sharp constant inequality of Poincaré type optimal control

1 Introduction

Denote by W 1 , 2 ( 1 , 1 ) the Sobolev space of all real-valued functions f ( ) that are absolutely continuous on the closed interval [ 1 , 1 ] and such that f ( ) L 2 ( 1 , 1 ) . Let m 1 be an integer. Denote by W ( 1 , 2 , m ) the space
W ( 1 ; 2 , m ) = { y ( ) W 1 , 2 ( 1 , 1 ) | 1 1 t k y ( t ) d t = 0 ( 0 k m 1 ) } .
(1.1)

Kalyabin considered in [1] the following problem.

Problem ( B x ) Fix x [ 1 , 1 ] , find the best constant B m ( x ) such that the following inequality holds:
| y ( x ) | B m ( x ) ( 1 1 | y ( t ) | 2 d t ) 1 2 , y ( ) W ( 1 , 2 , m ) .
(1.2)
It is proved in [1] that
B m 2 ( ± 1 ) = 2 m ( m + 2 ) , B 1 2 ( x ) = 1 + 3 x 2 6 , B 2 2 ( 0 ) = 1 6
(1.3)
and the extremal functions for the case x = 1 are
C ( ( m + 2 ) P m ( t ) + m P m + 1 ( t ) ) ,
(1.4)
where C is a constant and
P k ( t ) = 1 2 k k ! d k d t k ( t 2 1 ) k , k = 0 , 1 , 2 , ,
(1.5)
are the classical Legendre polynomials. For notational simplicity, we denote
p k ( t ) = 1 2 k k ! d k 2 d t k 2 ( t 2 1 ) k , k = 2 , 3 ,
(1.6)
and
q k ( x ) = 1 1 P k ( s ) | s x | d s 1 1 | P k ( s ) | 2 d s = { x 2 + 1 2 , if  k = 0 , x 3 3 x 2 , if  k = 1 , ( 2 k + 1 ) p k ( x ) , if  k 2 .
(1.7)

Many problems similar to Problem ( B x ) were studied; see, for example, [2, 3], and [4].

In this paper, we will solve Problem ( B x ) completely with the help of optimal control theory. Since the cases x = ± 1 were solved in [1], we mainly consider the cases of x ( 1 , 1 ) . We have the following.

Theorem 1.1 Assume m 1 and x ( 1 , 1 ) . Then, y ( ) W ( 1 , 2 , m ) is an extremal function to Problem ( B x ) if and only if
y ( t ) = C [ c ( x ) ( Q m + 1 ( t ) | t x | ) + 1 ] , t [ 1 , 1 ] ,
(1.8)
where C is a constant,
Q m ( t ) = A ( x ) + 1 k m 1 q k ( x ) P k ( t ) + α ( x ) P m ( t ) + β ( x ) P m + 1 ( t ) , t [ 1 , 1 ] ,
(1.9)
c ( x ) = 2 x 2 + 1 2 A ( x ) ,
(1.10)
and A ( x ) , α ( x ) , β ( x ) are characterized by
Q m ( 1 ) = 1 , Q m ( 1 ) = 1
(1.11)
and
Q m ( x ) = 0 .
(1.12)
The sharp constant of the inequality (1.2) is
B m ( x ) = 1 2 c ( x ) , x ( 1 , 1 ) .
(1.13)

More precisely, we have the following.

Corollary 1.2 Assume m = 1 and x ( 1 , 1 ) . Then
B 1 ( x ) = 3 x 2 + 1 6
(1.14)
and y ( ) W ( 1 , 2 , 1 ) is an extremal function to Problem ( B x ) if and only if
y ( t ) = C [ 3 3 x 2 + 1 ( t 2 x 2 2 | t x | ) + 1 ] , t [ 1 , 1 ] .
(1.15)
Corollary 1.3 Assume m = 2 and x ( 1 , 1 ) . Then
B 2 ( x ) = 8 21 x 2 + 30 x 4 5 x 6 48
(1.16)
and y ( ) W ( 1 , 2 , 2 ) is an extremal function to Problem ( B x ) if and only if
y ( t ) = C [ 4 + 33 x 2 30 x 4 + 5 x 6 + 8 P 2 ( t ) 2 ( x 3 3 x ) ( P 3 ( t ) 6 P 1 ( t ) ) 24 | t x | 8 21 x 2 + 30 x 4 5 x 6 + 1 ] .
(1.17)
Corollary 1.4 Assume m = 2 n + 1 , n 1 and x ( 1 , 1 ) . Then
B 2 n + 1 ( x ) = { ( x 2 1 ) 2 4 + 1 2 k = 2 2 n ( 2 k + 1 ) p k ( x ) P k ( x ) 1 2 ( n + 1 ) ( 2 n + 1 ) [ x 3 3 x 2 + 1 k n 1 ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] P 2 n + 1 ( x ) + 1 2 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] P 2 n + 2 ( x ) } 1 2 .
(1.18)
In particular,
B 3 ( x ) = 297 + 1 , 260 x 2 5 , 370 x 4 + 5 , 900 x 6 1 , 575 x 8 16 15 ,
(1.19)
B 5 ( x ) = 1 , 375 + 8 , 400 x 2 95 , 025 x 4 + 357 , 560 x 6 597 , 555 x 8 + 448 , 056 x 10 121 , 275 x 12 16 105 .
(1.20)
Corollary 1.5 Assume m = 2 n + 2 , n 1 and x ( 1 , 1 ) . Then
B 2 n + 2 ( x ) = { ( x 2 1 ) 2 4 + 1 2 k = 2 2 n + 1 ( 2 k + 1 ) p k ( x ) P k ( x ) + 1 2 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] P 2 n + 2 ( x ) 1 2 ( n + 2 ) ( 2 n + 3 ) [ x 3 3 x 2 + k = 1 n ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] × P 2 n + 3 ( x ) } 1 2 .
(1.21)
In particular,
B 4 ( x ) = 297 1 , 440 x 2 + 9 , 030 x 4 20 , 860 x 6 + 18 , 585 x 8 5 , 292 x 10 16 15 ,
(1.22)
B 6 ( x ) = 1 32 42 ( 2 , 200 15 , 225 x 2 + 211 , 050 x 4 1 , 162 , 455 x 6 + 3 , 017 , 700 x 8 3 , 977 , 127 x 10 + 2 , 562 , 714 x 12 637 , 065 x 14 ) 1 2 .
(1.23)

When x = 0 , we have the following.

Corollary 1.6 The following holds:
B 1 2 ( 0 ) = B 2 2 ( 0 ) = 1 6 , B 3 2 ( 0 ) = B 4 2 ( 0 ) = 99 1 , 280
(1.24)
and
B 2 n + 1 2 ( 0 ) = B 2 n + 2 2 ( 0 ) = 3 32 k = 2 n ( 4 k + 1 ) 2 4 k + 2 ( k + 1 ) ( 2 k 1 ) ( 2 k k ) 2 + ( 1 ) n ( 2 n + 2 n + 1 ) 2 2 n + 2 ( 2 n + 3 2 ) [ 7 16 k = 2 n ( 1 ) k k ( 4 k + 1 ) 2 2 k + 3 ( 2 k 1 ) ( 2 k + 2 k + 1 ) ] , n 2 .
(1.25)
In particular,
B 5 2 ( 0 ) = B 6 2 ( 0 ) = 275 5 , 376 , B 7 2 ( 0 ) = B 8 2 ( 0 ) = 45 , 325 1 , 179 , 648 .
(1.26)

2 Transmitting Problem ( B x ) to optimal control problem

We introduce the equivalent optimal control problem to Problem ( B x ). Let U = L 2 ( 1 , 1 ) . We define the following control system:
d d t ( y ( t ) w ( t ) z 0 ( t ) z 1 ( t ) z m 1 ( t ) ) = ( u ( t ) u ( t ) χ ( 1 , x ) ( t ) y ( t ) t y ( t ) t m 1 y ( t ) ) , t ( 1 , 1 )
(2.1)
and the state constraints
y ( 1 ) = w ( 1 ) , w ( 1 ) = 1 , z k ( ± 1 ) = 0 ( k = 0 , 1 , , m 1 ) .
(2.2)
Let
P a d = { ( Y ( ) , u ( ) ) ( W 1 , 2 ( 1 , 1 ) ) m + 2 × U | ( Y ( ) , u ( ) )  satisfying  ( 2.1 ) - ( 2.2 ) }
(2.3)
and
U a d = { u ( ) | ( Y ( ) , u ( ) ) P a d } ,
(2.4)
where
Y ( ) = ( y ( ) w ( ) z 0 ( ) z 1 ( ) z m 1 ( ) ) .
(2.5)

Our optimal control problem corresponding to Problem ( B x ) is as follows.

Problem ( C x ) Let x ( 1 , 1 ) . Find ( Y ¯ ( ) , u ¯ ( ) ) P a d such that
1 1 u ¯ 2 ( t ) d t = inf ( Y ( ) , u ( ) ) P a d 1 1 u 2 ( t ) d t .
(2.6)
It is obvious that ( Y ( ) , u ( ) ) y ( ) is a bijection from P a d to { f ( ) W ( 1 , 2 , m ) | f ( x ) = 1 } . Then one can easily see that
B m ( x ) = ( inf ( Y ( ) , u ( ) ) P a d 1 1 u 2 ( t ) d t ) 1 2 .
(2.7)

Therefore, we can solve Problem ( B x ) by solving Problem ( C x ).

3 Pontryagin’s maximum principle

We state Pontryagin’s maximum principle for optimal control problems. Symbols in this section will have similar but probably different meanings from other sections. Thus we set this part as a separate section. We will state a result given in [5]. For simplicity, we only state it in a simple way. In other words, Lemma 3.1 below is a special case of Theorem 3.1 and Corollary 3.1 in Chapter V of [5].

Now, let T > t 0 and U R m . A measurable function u ( ) defined on [ t 0 , T ] with range in U is said to be a control.

Let the function f ˆ = ( f 0 , f ) = ( f 0 , f 1 , , f n ) be an R n + 1 -valued vector function on [ t 0 , T ] × R n × R m . Assume that f ˆ is Borel measurable on t [ t 0 , T ] , continuous on ( y , u ) R n × R m and continuously differentiable on y R n .

If y ( ) is an absolutely function on [ t 0 , T ] with range in R n such that
d y ( t ) d t = f ( t , y ( t ) , u ( t ) ) , a.e. on  [ t 0 , T ] ,

then y ( ) is called a state/trajectory corresponding to u ( ) .

Let Ω R n × R n be a C 1 manifold of dimensional k, where 0 k 2 n . We say ( y ( ) , u ( ) ) is an admissible pair if
  1. (i)

    u ( ) is a control,

     
  2. (ii)

    y ( ) is a state corresponding to u ( ) ,

     
  3. (iii)

    ( y ( t 0 ) , y ( T ) ) Ω .

     

Denote by P a d the set of all admissible pairs. The set U a d = { u ( ) | ( Y ( ) , u ( ) ) P a d } is called the set of admissible controls.

If an admissible pair ( y ¯ ( ) , u ¯ ( ) ) satisfies
J ( y ¯ ( ) , u ¯ ( ) ) = inf ( y ( ) , u ( ) ) P a d J ( y ( ) , u ( ) ) ,
then it is called an optimal pair, where
J ( y ( ) , u ( ) ) : = 0 T f 0 ( t , y ( t ) , u ( t ) ) d t .
Assume that for each compact Y R n and admissible control u ( ) , there exists a function μ ( ) = μ ( ; Y , u ( ) ) such that for almost all t [ t 0 , T ] and all y Y ,
| f ˆ ( t , y , u ( t ) ) | μ ( t ) , | f ˆ y ( t , y , u ( t ) ) | μ ( t ) .
(3.1)

We have the following.

Lemma 3.1 Let assumptions listed in this section hold. Let ( y ¯ ( ) , u ¯ ( ) ) be an optimal pair. Then there exists a constant λ 0 0 and an absolutely continuous vector function λ = ( λ 1 , λ 2 , , λ n ) defined on [ t 0 , T ] such that the following hold:
  1. (i)

    The vector ( λ 0 , λ ( t ) ) is never zero on [ t 0 , T ] .

     
  2. (ii)
    For a.e. t [ t 0 , T ] ,
    λ ( t ) = f y ( t , y ¯ ( t ) , u ¯ ( t ) ) λ ( t ) λ 0 f y 0 t , y ¯ ( t ) , u ¯ ( t ) ) ,
     
where
f y = ( f 1 y 1 f 2 y 1 f n y 1 f 1 y 2 f 2 y 2 f n y 2 f 1 y n f 2 y n f n y n ) .
  1. (iii)
    The pointwise maximum condition holds: for almost all t [ t 0 , T ] and all u U ,
    λ 0 f 0 ( t , y ¯ ( t ) , u ¯ ( t ) ) + λ ( t ) , f ( t , y ¯ ( t ) , u ¯ ( t ) ) λ 0 f 0 ( t , y ¯ ( t ) , u ¯ ( t ) ) + λ ( t ) , f ( t , y ¯ ( t ) , u ) .
     
  2. (iv)

    The transversality condition holds: if the mapping t f ˆ ( t , y ¯ ( t ) , u ¯ ( t ) ) is continuous at t = t 0 and t = T , then ( λ ( t 0 ) , λ ( T ) ) is orthogonal to Ω.

     

4 Proof of Theorem 1.1

We give the following lemma first.

Lemma 4.1 Let n 1 , c R , x ( 1 , 1 ) , Q ( ) is an ( n + 1 ) th degree polynomial satisfying
Q ( x ) = 0 , Q ( 1 ) = 1 , Q ( 1 ) = 1
(4.1)
and
1 1 t k [ c ( Q ( t ) | t x | ) + 1 ] d t = 0 , k = 0 , 1 , , n 1 .
(4.2)
Then
1 1 [ t ( Q ( t ) | t x | ) ] 2 d t = 2 c .
(4.3)
Proof Noting that Q ( ) is an ( n 1 ) th degree polynomial, by (4.2), we have
1 1 Q ( t ) [ c ( Q ( t ) | t x | ) + 1 ] d t = 0 .
(4.4)
Therefore,
1 1 [ t ( Q ( t ) | t x | ) ] 2 d t = 1 x ( Q ( t ) + 1 ) 2 d t + x 1 ( Q ( t ) 1 ) 2 d t = 1 1 ( Q ( t ) ) 2 d t + 2 + 2 1 x Q ( t ) d t 2 x 1 Q ( t ) d t = Q ( 1 ) Q ( 1 ) Q ( 1 ) Q ( 1 ) 1 1 Q ( t ) Q ( t ) d t + 2 + 4 Q ( x ) 2 Q ( 1 ) 2 Q ( 1 ) = 1 1 Q ( t ) ( 1 c | t x | ) d t + 2 Q ( 1 ) Q ( 1 ) = 2 c + 1 x Q ( t ) ( t x ) d t + x 1 Q ( t ) ( x t ) d t + 2 Q ( 1 ) Q ( 1 ) = 2 c ( 1 + x ) 1 x Q ( t ) d t ( 1 x ) + x 1 Q ( t ) d t + 2 Q ( 1 ) Q ( 1 ) = 2 c .
(4.5)

 □

Now, we list some properties of Legendre polynomials. We can easily get
P k ( 0 ) = { 0 , if  k = 2 n + 1 , n 0 , 1 , if  k = 0 , ( 1 ) n ( 2 n 1 ) ! ! ( 2 n ) ! ! , if  k = 2 n , n 1 ,
(4.6)
p k ( 0 ) = { 0 , if  k = 2 n + 1 , n 1 , 1 8 , if  k = 2 , ( 1 ) n 1 ( 2 n 3 ) ! ! ( 2 n + 2 ) ! ! , if  k = 2 n , n 2 ,
(4.7)
P k ( 1 ) = ( 1 ) k , P k ( 1 ) = 1 ,
(4.8)
P k ( 1 ) = ( 1 ) k 1 k ( k + 1 ) 2 , P k ( 1 ) = k ( k + 1 ) 2 ,
(4.9)
1 1 | P k ( t ) | 2 d t = 2 2 k + 1 ,
(4.10)
and
1 1 ( P k ( t ) ) 2 d t = P k ( 1 ) P k ( 1 ) P k ( 1 ) P k ( 1 ) = k ( k + 1 ) ,
(4.11)
1 1 P k ( t ) P k + 1 ( t ) d t = P k ( 1 ) P k + 1 ( 1 ) P k ( 1 ) P k + 1 ( 1 ) = 0 ,
(4.12)
1 1 P k ( t ) | t x | d t = { x 2 + 1 , if  k = 0 , 1 3 x 3 x , if  k = 1 , 2 p k ( x ) , if  k 2 .
(4.13)

We turn to the proof of Theorem 1.1.

Proof of Theorem 1.1 I. Existence of optimal pair. One can prove directly that the sharp constant B m ( x ) is attainable, i.e., there is a nontrivial y ¯ ( ) W ( 1 , 2 , m ) such that
| y ¯ ( x ) | = B m ( x ) ( 1 1 | y ( t ) | 2 d t ) 1 2 .
(4.14)

Now, we give an optimal control version of this fact.

Let ( Y j ( ) , u j ( ) ) P a d be a minimizing sequence of Problem ( C x ). That is
lim j + 1 1 u j 2 ( t ) d t = inf ( Y ( ) , u ( ) ) P a d 1 1 u 2 ( t ) d t .
(4.15)
Then u j ( ) is bounded in L 2 ( 1 , 1 ) . That is
u j ( ) L 2 ( 1 , 1 ) M
for some constant M > 0 . Denote Y j ( ) ( y j ( ) , w j ( ) , z 0 j ( ) , , z m 1 j ( ) ) . Then, by the state equation (2.1) and the constraints (2.2), we have
1 1 y j ( t ) d t = z 0 j ( 1 ) z 0 j ( 1 ) = 0 .
Then (2.1)-(2.2), and Poincaré’s inequality imply
y j ( ) W 1 , 2 ( 1 , 1 ) C 1 , w j ( ) W 1 , 2 ( 1 , 1 ) C 1 , j 1 ,
(4.16)
and consequently
z k j ( ) W 1 , 2 ( 1 , 1 ) C 2 , k = 0 , 1 , , m 1 ; j 1
(4.17)

for some constant C 1 , C 2 . That is, Y j ( ) is bounded in ( W 1 , 2 ( 1 , 1 ) ) m + 2 . Then, by Sobolev’s imbedding theorem, Y j ( ) is bounded and equicontinuous in ( C [ 1 , 1 ] ) m + 2 .

Thus, Eberlein-Shmulyan’s theorem and Arzelá-Ascoli’s theorem (see Chapter V, Appendix 4 and Chapter III, Section 3 in [6], for example), we can suppose that
u j ( ) u ¯ ( ) , weakly in  L 2 ( 1 , 1 )
(4.18)
and
Y j ( ) Y ¯ ( ) , weakly in  ( W 1 , 2 ( 1 , 1 ) ) m + 2 ,  strongly in  ( C [ 1 , 1 ] ) m + 2
(4.19)
for some u ¯ ( ) L 2 ( 1 , 1 ) and Y ¯ ( ) ( W 1 , 2 ( 1 , 1 ) ) m + 2 . One can easily see that ( Y ¯ ( ) , u ¯ ( ) ) satisfies (2.1) and (2.2). Thus, ( Y ¯ ( ) , u ¯ ( ) ) P a d . Moreover,
1 1 u ¯ 2 ( t ) d t lim j + 1 1 u j 2 ( t ) d t .
(4.20)

Therefore ( Y ¯ ( ) , u ¯ ( ) ) is a solution to Problem ( C x ). We call it an optimal pair of Problem ( C x ).

II. Pontryagin’s maximum principle for the optimal pair. We now apply Lemma 4.1 - Pontryagin’s maximum principle to Problem ( C x ). We can easily verify that all the conditions posed in Section 3 hold. For example, conditions on state constraints and the local existence of a dominating integrable function (see (3.1)) hold. More precisely, let
Ω = { ( s , s ) | s R } × { 0 } m × R × { 1 } × { 0 } m .

Then Ω is a C 1 manifold of dimensional 2. While the state constraints (2.2) is equivalent to ( Y ( 1 ) , Y ( 1 ) ) Ω .

On the other hand, for any u ( ) L 2 ( 1 , 1 ) and | Y | R , if we choose μ ( t ) = 2 | u ( t ) | + m R , then the condition (4.1) corresponding to (2.1) holds.

Now, by Lemma 4.1, the optimal pair ( Y ¯ ( ) , u ¯ ( ) ) satisfies the following Pontryagin maximum principle: there exists a φ 0 0 and a solution to the following adjoint equation:
d d t ( φ ( t ) ζ ( t ) ψ 0 ( t ) ψ 1 ( t ) ψ m 1 ( t ) ) = ( j = 0 m 1 t j ψ j ( t ) 0 0 0 0 ) , t ( 1 , 1 )
(4.21)
such that the following conditions hold:
  1. (i)
    we have the following non-trivial condition:
    ( φ 0 , φ ( ) , ζ ( ) , ψ 0 ( ) , , ψ m 1 ( ) ) 0 ,
    (4.22)
     
  2. (ii)
    the maximum condition:
    φ 0 u ¯ 2 ( t ) + ( φ ( t ) + ζ ( t ) χ ( 1 , x ) ( t ) ) u ¯ ( t ) = max u R [ φ 0 u 2 + ( φ ( t ) + ζ ( t ) χ ( 1 , x ) ( t ) ) u ] , a.e.  t ( 1 , 1 ) ,
    (4.23)
     
  3. (iii)
    the transversality condition
    φ ( 1 ) + ζ ( 1 ) = 0 , φ ( 1 ) = 0 .
    (4.24)
     
III. Simplification. By (4.21), ζ ( ) ζ , ψ 0 ( ) ψ 0 , , ψ m 1 ( ) ψ m 1 are constants and
φ ( t ) = j = 0 m 1 ψ j t j , t ( 1 , 1 ) .
(4.25)

Then φ ( ) is an m th degree polynomial.

By (4.23), we have
2 φ 0 u ¯ ( t ) + φ ( t ) + ζ χ ( 1 , x ) = 0 , a.e.  t ( 1 , 1 ) .
(4.26)
If φ 0 = 0 , we get
φ ( t ) = { ζ , t ( 1 , x ) , 0 , t [ x , 1 ) , a.e.  t ( 1 , 1 ) .
Therefore, since φ ( ) is a polynomial, we have the following:
φ ( ) 0 , ζ = 0 .
(4.27)
Consequently, by (4.25),
ψ 0 = ψ 1 = = ψ m 1 = 0 .
(4.28)
This contradicts the non-trivial condition (4.22). Therefore, we must have φ 0 < 0 . Without loss of generality, we can suppose that φ 0 = 1 2 . Then it follows from (4.26) that
u ¯ ( t ) = φ ( t ) + ζ χ ( 1 , x ) ( t ) , t ( 1 , 1 ) .
(4.29)
Combining the above with (4.24), we see that the corresponding function y ¯ ( ) is continuously differentiable on [ 1 , x ) ( x , 1 ] and
y ¯ ( ± 1 ) = u ¯ ( ± 1 ) = 0 .
(4.30)
Moreover, y ¯ ( ) can be expressed as
y ¯ ( t ) = c ( x ) | t x | + Q ˜ m ( t ) , t [ 1 , 1 ] ,
(4.31)

where c c ( x ) = ζ 2 and Q ˜ m ( ) is an ( m + 1 ) th degree polynomial.

We claim that c 0 . Otherwise, c = 0 . Then it follows from (4.31) and
1 1 t k y ¯ ( t ) d t = 0 , k = 0 , 1 , 2 , , m 1 ,
(4.32)
that
Q ˜ m ( t ) = c m P m ( t ) + c m + 1 P m + 1 ( t ) , k = 0 , 1 , 2 , , m 1
(4.33)

for some constant c m , c m + 1 .

Then (4.30) and (4.9) imply c m = c m + 1 = 0 . This contradicts the nontrivial condition. Therefore c 0 and we can rewrite y ¯ ( ) as
y ¯ ( t ) = c ( x ) ( Q m ( t ) | t x | ) + 1 , t [ 1 , 1 ] ,
(4.34)
where Q m ( ) is an ( m + 1 ) th degree polynomial such that
Q m ( x ) = 0 , Q m ( 1 ) = 1 , Q m ( 1 ) = 1 .
(4.35)
IV. Conclusion. By (4.32),
1 1 Q m ( t ) P k ( t ) d t = 1 1 P k ( t ) | x t | d t , 1 k m 1 .
(4.36)
Thus we see that
Q m ( t ) = A ( x ) + 1 k m 1 q k ( x ) P k ( t ) + α ( x ) P m ( t ) + β ( x ) P m + 1 ( t ) ,
(4.37)
where q k ( x ) is defined by (1.7). Moreover, by (4.35), we can determine A ( x ) , α ( x ) and β ( x ) . Finally, using (4.32) again, we get
c ( x ) = 2 x 2 + 1 2 A ( x ) .
(4.38)
By Lemma 4.1, we have the following:
1 1 | u ¯ ( t ) | 2 d t = 1 1 | y ¯ ( t ) | 2 d t = 2 c ( x ) .
(4.39)

Then Theorem 1.1 follows from (2.7). □

Remark 4.1 If x = 1 , instead of (4.34)-(4.35), we get
y ¯ ( t ) = Q m ( t ) , t [ 1 , 1 ]
(4.40)
with
Q m ( 1 ) = 1 , Q m ( 1 ) = 0
(4.41)
and
Q m ( t ) = α P m ( t ) + β P m + 1 ( t ) , t [ 1 , 1 ] .
(4.42)

The above equations imply the results in [1] for x = ± 1 .

On the other hand, since B m ( x ) is obviously continuous respect to x [ 1 , 1 ] , we can certainly get B m ( ± 1 ) from Theorem 1.1.

5 Results for some special cases

We prove Corollaries 1.2-1.6 in this section.

Proof of Corollary 1.2 By (1.9), we have
Q 1 ( t ) = A ( x ) + α ( x ) P 1 ( t ) + β ( x ) P 2 ( t ) .
(5.1)
Thus (1.11) and (4.9) imply α ( x ) = 0 , β ( x ) = 1 3 . Then (1.12) implies
A ( x ) = 1 3 P 2 ( x ) = 1 3 x 2 6 .
(5.2)
Consequently,
Q 1 ( t ) = 1 3 ( P 2 ( t ) P 2 ( x ) ) = t 2 x 2 2
(5.3)
and
c ( x ) 2 x 2 + 1 2 A ( x ) = 3 3 x 2 + 1 .
(5.4)
Therefore, the extremal functions to Problem ( B x ) are C y ¯ ( ) with
y ¯ ( x ) = 3 3 x 2 + 1 ( t 2 x 2 2 | t x | ) + 1 ,
(5.5)
while
B 1 ( x ) = 1 2 c ( x ) = 3 x 2 + 1 6 .
(5.6)

 □

Proof of Corollary 1.3 By (1.7) and (1.9), we have
Q 2 ( t ) = A ( x ) + x 3 3 x 2 P 1 ( t ) + α ( x ) P 2 ( t ) + β ( x ) P 3 ( t ) .
(5.7)
Then it follows easily from (1.11) that
α ( x ) = 1 3 , β ( x ) = x 3 3 x 12 .
(5.8)
Then, by (1.12),
A ( x ) = 4 + 33 x 2 30 x 4 + 5 x 6 24 .
(5.9)
Thus
c ( x ) = 2 x 2 + 1 2 A ( x ) = 24 8 21 x 2 + 30 x 4 5 x 6 .
(5.10)
Therefore
B 2 ( x ) = 1 2 c ( x ) = 8 21 x 2 + 30 x 4 5 x 6 48
(5.11)
and the extremal functions to Problem ( B x ) are
C ( 4 + 33 x 2 30 x 4 + 5 x 6 + 8 P 2 ( t ) 2 ( x 3 3 x ) ( P 3 ( t ) 6 P 1 ( t ) ) 24 | t x | 8 21 x 2 + 30 x 4 5 x 6 + 1 ) .
(5.12)

 □

Proof of Corollary 1.4 By (1.7) and (1.9), we have
Q 2 n + 1 ( t ) = A ( x ) + x 3 3 x 2 P 1 ( t ) + k = 2 2 n ( 2 k + 1 ) p k ( x ) P k ( t ) + α ( x ) P 2 n + 1 ( t ) + β ( x ) P 2 n + 2 ( t ) .
(5.13)
Then by (1.11)-(1.12), and (1.10),
α ( x ) = 1 ( n + 1 ) ( 2 n + 1 ) [ x 3 3 x 2 + 1 k n 1 ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] ,
(5.14)
β ( x ) = 1 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] ,
(5.15)
A ( x ) = x 3 3 x 2 P 1 ( x ) k = 2 2 n ( 2 k + 1 ) p k ( x ) P k ( x ) A ( x ) = + 1 ( n + 1 ) ( 2 n + 1 ) [ x 3 3 x 2 + 1 k n 1 ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] P 2 n + 1 ( x ) A ( x ) = 1 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] P 2 n + 2 ( x ) ,
(5.16)
1 2 c ( x ) = x 2 + 1 4 A ( x ) 2 = ( x 2 1 ) 2 4 + 1 2 k = 2 2 n ( 2 k + 1 ) p k ( x ) P k ( x ) 1 2 c ( x ) = 1 2 ( n + 1 ) ( 2 n + 1 ) [ x 3 3 x 2 + 1 k n 1 ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] P 2 n + 1 ( x ) 1 2 c ( x ) = + 1 2 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] P 2 n + 2 ( x ) .
(5.17)

Finally, (1.19) and (1.20) follow from direct calculations. We get the proof. □

Proof of Corollary 1.5 By (1.7) and (1.9), we have
Q 2 n + 2 ( t ) = A ( x ) + x 3 3 x 2 P 1 ( t ) + k = 2 2 n + 1 ( 2 k + 1 ) p k ( x ) P k ( t ) + α ( x ) P 2 n + 2 ( t ) + β ( x ) P 2 n + 3 ( t ) .
(5.18)
Then by (1.11)-(1.12), and (1.10),
α ( x ) = 1 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] ,
(5.19)
β ( x ) = 1 ( n + 2 ) ( 2 n + 3 ) [ x 3 3 x 2 + k = 1 n ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] ,
(5.20)
A ( x ) = x 3 3 x 2 P 1 ( x ) k = 2 2 n + 1 ( 2 k + 1 ) p k ( x ) P k ( x ) A ( x ) = 1 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] P 2 n + 2 ( x ) A ( x ) = + 1 ( n + 2 ) ( 2 n + 3 ) [ x 3 3 x 2 + k = 1 n ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] A ( x ) = × P 2 n + 3 ( x ) ,
(5.21)
1 2 c ( x ) = x 2 + 1 4 A ( x ) 2 1 2 c ( x ) = ( x 2 1 ) 2 4 + 1 2 k = 2 2 n + 1 ( 2 k + 1 ) p k ( x ) P k ( x ) 1 2 c ( x ) = + 1 2 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( x ) ] P 2 n + 2 ( x ) 1 2 c ( x ) = 1 2 ( n + 2 ) ( 2 n + 3 ) [ x 3 3 x 2 + k = 1 n ( k + 1 ) ( 2 k + 1 ) ( 4 k + 3 ) p 2 k + 1 ( x ) ] 1 2 c ( x ) = × P 2 n + 3 ( x ) .
(5.22)

Finally, (1.21) and (1.22) follow from direct calculations. We get the proof. □

Proof of Corollary 1.6 First, we get (1.24) from (1.14), (1.16), (1.19), and (1.22).

By (4.7), p 2 k + 1 ( 0 ) = 0 ( k = 1 , 2 , ). Thus, if n 2 , we get from (1.18) and (1.21)
B 2 n + 1 2 ( 0 ) = B 2 n + 2 2 ( 0 ) = 1 4 + 1 2 k = 1 n ( 4 k + 1 ) p 2 k ( 0 ) P 2 k ( 0 ) + 1 2 ( n + 1 ) ( 2 n + 3 ) [ 1 k = 1 n k ( 2 k + 1 ) ( 4 k + 1 ) p 2 k ( 0 ) ] P 2 n + 2 ( 0 ) .
(5.23)
Moreover, using (4.6)-(4.7), we get (1.25):
B 2 n + 1 2 ( 0 ) = B 2 n + 2 2 ( 0 ) = 1 4 + 1 2 × 5 8 × ( 1 2 ) 1 2 k = 2 n ( 4 k + 1 ) ( 2 k 3 ) ! ! ( 2 k + 2 ) ! ! ( 2 k 1 ) ! ! ( 2 k ) ! ! ( 1 ) n 2 ( n + 1 ) ( 2 n + 3 ) [ 1 15 8 + k = 2 n ( 1 ) k k ( 2 k + 1 ) ( 4 k + 1 ) ( 2 k 3 ) ! ! ( 2 k + 2 ) ! ! ] ( 2 n + 1 ) ! ! ( 2 n + 2 ) ! ! = 3 32 k = 2 n ( 4 k + 1 ) 2 4 k + 2 ( k + 1 ) ( 2 k 1 ) ( 2 k k ) 2 + ( 1 ) n ( 2 n + 2 n + 1 ) 2 2 n + 2 ( 2 n + 3 2 ) [ 7 16 k = 2 n ( 1 ) k k ( 4 k + 1 ) 2 2 k + 3 ( 2 k 1 ) ( 2 k + 2 k + 1 ) ] , n 2 .

Now (1.26) follows directly from (1.25). □

Declarations

Acknowledgements

This work was supported in part by 973 Program (No. 2011CB808002) and NSFC (No. 11371104).

Authors’ Affiliations

(1)
School of Mathematical Sciences, and LMNS, Fudan University

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© Lou; licensee Springer. 2014

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